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Proof step 24 from proof max-ins (previous step)

a0 < a, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) = max(x0), x0 = [], max(x0) < a0, max(x0) < a

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 25:
a0 < a, max(ins≤(a, x0)) < a0, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) = max(x0), x0 = [], max(x0) < a, max(x0) < a0
Proof step 27:
a0 < a, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) = max(x0), x0 = [], max(x0) < a, max(x0) < a0, max(ins≤(a, x0)) < a0

Rule argument

fma: max(ins≤(a, x0)) < a0

Additional information

Used formulas: max(ins≤(a, x0)) < a0